1955 AMC 12 Problem 30

Attempt Problem 30 of the 1955 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1955 AMC 12 solutions, or check the answer key.

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30.

Each of the equations 3x2−2=25,3x^2-2=25, (2x−1)2=(x−1)2,(2x-1)^2=(x-1)^2, x2−7=x−1\sqrt{x^2-7}=\sqrt{x-1} has:
two integral roots
no root greater than 33
no root zero
only one root
one negative root and one positive root
Answer: B
Concepts:equation solvingradical domaincommon property
Small hint:
Solve each equation separately and compare the properties of their roots
Big hint:
For the radical equation, reject candidates that make either radicand negative
Solution:
The first equation gives x=±3.x=\pm3. Factoring the difference of squares in the second gives [(2x−1)−(x−1)]⋅[(2x−1)+(x−1)]=x(3x−2)=0, \begin{aligned} &[(2x-1)-(x-1)]\\ &\quad\cdot[(2x-1)+(x-1)]\\ &\qquad=x(3x-2)=0, \end{aligned} so x=0x=0 or 23.\frac{2}{3}. Squaring the third gives x2−x−6=0,x^2-x-6=0, with candidates 33 and −2;-2; only 33 satisfies the real-domain restrictions. Every root obtained is at most 3.3.
Thus, each equation has no root greater than 3,3, and the correct answer is B.
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